1. **State the problem:** We want to find the sample size $n$ needed for a hypothesis test of $H_0: \mu = 10$ against an alternative where the true mean is $11.2$, with a significance level $\alpha = 0.05$ and power at least $80\%$ (i.e., probability of rejecting $H_0$ when $\mu = 11.2$ is at least 0.8). 2. **Set up the test:** Assume the population standard deviation $\sigma$ is known or estimated. The test statistic for the sample mean $\bar{X}$ is $$Z = \frac{\bar{X} - 10}{\sigma / \sqrt{n}}$$ 3. **Significance level:** For a one-sided test at $\alpha = 0.05$, the critical value is $$z_{\alpha} = 1.645$$ We reject $H_0$ if $$Z > 1.645$$ 4. **Power condition:** The power is the probability of rejecting $H_0$ when $\mu = 11.2$. The distribution of $Z$ under $\mu = 11.2$ is $$Z = \frac{\bar{X} - 10}{\sigma / \sqrt{n}} = \frac{(\bar{X} - 11.2) + 1.2}{\sigma / \sqrt{n}}$$ Since $\bar{X}$ is centered at $11.2$, the mean of $Z$ under $\mu=11.2$ is $$\frac{11.2 - 10}{\sigma / \sqrt{n}} = \frac{1.2 \sqrt{n}}{\sigma}$$ 5. **Power formula:** The power is $$P\left(Z > 1.645 \mid \mu=11.2\right) = P\left(Z' > 1.645 - \frac{1.2 \sqrt{n}}{\sigma}\right)$$ where $Z' \sim N(0,1)$. We want this probability to be at least 0.8, so $$P\left(Z' > 1.645 - \frac{1.2 \sqrt{n}}{\sigma}\right) \geq 0.8$$ 6. **Find the quantile:** The $0.2$ quantile of the standard normal (since $P(Z' > z) = 0.8$ means $z$ is the 20th percentile) is $$z_{0.2} = -0.8416$$ So, $$1.645 - \frac{1.2 \sqrt{n}}{\sigma} = -0.8416$$ 7. **Solve for $n$:** $$\frac{1.2 \sqrt{n}}{\sigma} = 1.645 + 0.8416 = 2.4866$$ $$\sqrt{n} = \frac{2.4866 \sigma}{1.2}$$ $$n = \left(\frac{2.4866 \sigma}{1.2}\right)^2 = \frac{(2.4866)^2 \sigma^2}{1.44}$$ 8. **Final answer:** The required sample size depends on $\sigma$. If $\sigma$ is known, plug in its value to compute $n$. For example, if $\sigma = 1$, then $$n = \frac{(2.4866)^2}{1.44} \approx \frac{6.187}{1.44} \approx 4.3$$ So at least 5 signals need to be sent. **Summary:** $$n = \left(\frac{z_{\alpha} + z_{\beta}}{\frac{\mu_1 - \mu_0}{\sigma}}\right)^2$$ where $z_{\alpha} = 1.645$, $z_{\beta} = 0.8416$, $\mu_1 = 11.2$, $\mu_0 = 10$. This formula gives the minimum sample size to achieve the desired power and significance level.